Name …………………………………………….……… ADM/NUMBER…………..
231/ 1 Candidate’s Signature…………..
BIOLOGY
Paper 1 Date …………………………..
(Theory)
Time :2 hours
MOCKS 1 2023
Kenya Certificate of Secondary Education
Instructions to candidates
Write your name and class in the spaces provided above.
Append your signature and write the date of examination in the spaces provided above.
Spelling errors especially of biological terms shall be penalized
Candidates should answer the questions in English.
Answer ALL questions in the spaces provided.
For Examiner’s Use Only
|
Question |
Maximum Score |
Candidate’s Score |
|
1 – 29 |
80
|
|
This paper consists of 12 printed pages.
Candidates should check the question paper to ascertain that
all the pages are printed as indicated and no questions are missing.
- Name the group of organisms that are found on the boarder of living and non-living organisms [1mark]
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- State two features characteristic of fruits dispersed by animals [2marks]
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- Define each of the following biological phenomena
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- Double fertilisation in angiosperms [1 mark]
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- Name the disease condition caused by deficiency of each of the following
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- State the function of the mitochondrial cristae [1 mark]
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- Give thedifference between transpiration and guttation [2 marks]
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- Name two enzymes in the human digestive system which are secreted in an inactive form
[2 marks]
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- Outline the function of the cilia in the mammalian fallopian tube [1 mark]
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- Name the blood vessel that supplies blood to the
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- Explain why when placed in fresh water Entamoebahistolyticadoes not burst [2 marks]
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- (a) Name the taxonomic class of woodlice [1 mark]
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- Name two other organisms in the taxonomic class in (a) above [2 marks]
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- State two features characteristic of organisms in the taxonomic class in (a)(i) above
[2 marks]
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- To estimate the population size of grasshoppers in the 5km2 field behind the Matope school farm, a group of students caught 100 grasshoppers on the first day. They marked them and released them back into the field. 48 hours later, the students went back to the field and caught 80 grasshoppers. Of these, 40 were found to have been marked
- Suggest a suitable method used to mark the grasshoppers [1 mark]
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- Calculate the population density of grasshoppers in the field [3 marks]
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- The diagram below shows a specialized plant cell
- (i) Identify the cell [1 mark]
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- State one feature observable in the diagram above that adapts the cell to its function
[1 mark]
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- Name the part labelled F [1 mark]
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- Shown below is a diagram of a neurone
- (i) Identify the type of neurone in the diagram above [1 mark]
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- Give one reason for your answer in (a)(i) above [1 mark]
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- Draw an arrow, alongside the diagram above, to indicate the direction of flow of a nerve impulse in the neurone [1 mark]
- Study the equation below which represents a reaction which takes place in the mammalian body
Enzyme X
CO2(g)+ H2O(l)H2CO3(aq)
- Where in the mammalian body does the reaction above take place? [1 mark]
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- Name the enzyme X [1 mark]
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- The diagram below shows a specialized cell from a human being
- Name the part labelled P [1 mark]
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- Give one adaptive feature of the organelle labelled T [1 mark]
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- State the function of the part labelled Q [1 mark]
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- The apparatus shown in the diagram below may be used to demonstrate aerobic respiration. Air is drawn through the apparatus by attaching it to a vacuum pump at the point labelled X. Sodium hydroxide solution is placed in flask 1 to remove carbon(IV)oxide
1 2 small animal 3
- Why was it necessary to remove carbon(IV)oxide? [1 mark]
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- Calcium hydroxide solution is put in the jars labelled 2 and 3. Explain why [2 marks]
Jar 2………………………………………………………………………………………………………
Jar 3………………………………………………………………………………………………………
- Suggest a suitable control for this experiment [1 mark]
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- Illustrated below is the movement of material in a certain physiological process
- (i) Name the physiological process illustrated above [1 mark]
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- Give two examples of applicability of the process named in (a)(i) above in plants
[2 marks]
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- State two ways by which the movement of the dye molecules would be slowed down
[2 marks]
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- The diagrams below represent an experimental set up to investigate a certain biological phenomenon
START OF EXPERIMENT
END OF EXPERIMENT
- (i) What was being investigated in the experiment? [1 mark]
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- Explain your answer in (a)(i) above [3 marks]
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- What was the role of flask II in the experiment? [1 mark]
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- Outline two adaptive features of guard cells [2 marks]
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- The scientific name of the European wildcat, a nocturnal mammal that inhabits forests of Western, Southern, Central and Eastern Europe, is Felissilvestris. Write down
- the scientific name of the European wildcat [1 mark]
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- the species name of the European wildcat [1 mark]
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- Name the hormone, in man, responsible for each of the following
- Stimulates secretion of bile by hepatocytes [1 mark]
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- Stimulates release of bile juice from gall bladder into the duodenum [1 mark]
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- The following text messages on a cellular phone represent gene mutations
| |
Intended message |
Actual message |
| I |
Metereis a top school |
Metre is a top school |
| II |
The microscope is my tool |
The microscope is my loot |
Identify the type of gene mutation represented in each case
I………………………………………………………………………………………………… [1 mark]
II……………………………………………………………………………………………….. [1 mark]
- (a) Define omnivores [1 mark]
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- Name two mammals that are omnivores [2 marks]
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- (a) Two alleles in heterozygous state can be fully expressed phenotypically in an organism e.g.
the alleles for black and white skin colour in guinea pigs (Caviaporcellus). Give the term used
todescribe this phenomenon [1 mark]
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- Give one example of a trait in human beings where the condition whose term is named in (a) above expresses itself [1 mark]
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- Shown below are diagrams of the same mammalian blood cell
Surface view Cross section
State two morphological features of cell represented in the diagram above [2 marks]
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- The table below shows the effect of wind, still air and stomatal opening on the rate of transpiration of a plant in milligrams of water lost per hour dm2. Study the table and answer the following questions
| Stomatal opening (μm) |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
| Windy |
40 |
63 |
74 |
86 |
94 |
110 |
124 |
| Still air |
0 |
6 |
12 |
19 |
23 |
27 |
30 |
(a) (i) Compare the rates of transpiration in windy and still air conditions [1mark]
(ii) Explain your observation in a(i) above [2marks]
(b) How does stomatal opening affect transpiration rate? [1mark]
- The relationship between oxygen concentration, potassium uptake and sugar consumption in isolated barley roots was determined. The loss of sugar and potassium uptake are in arbitrary units
| |
Percentage oxygen in aeration stream |
| 0 |
5 |
10 |
15 |
20 |
100 |
| Sugar loss |
15 |
20 |
42 |
45 |
45 |
48 |
| Potassium gain |
5 |
55 |
70 |
73 |
75 |
70 |
- Account for the sugar loss and potassium gain at between 5% and 20% oxygen concentration [3marks]
- State two ways in which you can stop the above process from taking place [2marks]
- Illustrated in the diagrams below is the position of chloroplasts (shown as dark structures, ) in a tropical plant species, at two different times of the day
I II
At 6am At 2pm
- Identify the tissue labelled K [1 mark]
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- Name the response shown by the chloroplasts in diagram I [1 mark]
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- Outline the importance of the orientation of the chloroplasts as illustrated in diagram II
[2 marks]
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NAME:………………………………………………………………………………………….…………ADM/NO. ………………………….
SCHOOL:…………………………………………………………..………………SIGNATURE………………………DATE…………….…
231/2
BIOLOGY
PAPER 2
(Theory)
JULY 2023
TIME: 2HOURS
MOCKS 1 2023
Kenya Certificate of Secondary Education
INSTRUCTIONS TO CANDIDATES
- Write your name and Admission number in the spaces provided.
- Sign and write the date.
- This paper consists of two sections.A and B.
- Answer ALL the questions in section A in the spaces provided.
- In section B, answer question 6 (compulsory) and either question 7 or 8 in the spaces provided.
For examiners use only:
| Section |
Question |
Maximum score |
Candidates score |
| A |
1 |
8 |
|
| 2 |
8 |
|
| 3 |
8 |
|
| 4 |
8 |
|
| 5 |
8 |
|
| B |
6 |
20 |
|
| 7 |
20 |
|
| 8 |
20 |
|
| TOTAL SCORE |
80 |
|
SECTION A ( 40 MARKS)
- Below is a set up that was used to investigate a certain process in plants
- State what the above apparatus can be used to measure
- Directly [1mark]
- Indirectly [1mark]
- i) Give two precautions that should be taken when setting up the experiment[2marks]
(ii) State a reason for each precautions stated in b(i) above [2marks]
(c) List two structural factors that affect the process under investigation [2marks]
- A Covidiot at Nakuru National park wanted to estimate the population size of grasshoppers , 70 grasshoppers were trapped , marked and released. A week later , a second sample was captured. In this second sample , 27 had marks on their bodies while 13 did not have the marks.
- Calculate the estimated size of the grasshopper population [3marks]
- Explain why it is important that the samples contain as many grasshoppers as possible [1mark]
- Name an appropriate instrument that was used to capture the grasshoppers [1mark]
- Give 3 assumptions that must be made when using this method of population estimation [3marks]
- In an experiment, Drosophila melanogaster (fruit fly) with broad abdomens were crossed with those having narrow abdomens. All the F1 offspring from the crosses had broad abdomens :
- Using A to denote the genes for the abdomen size, give the genotypes of the parents[2marks]
- If 150 fruit flies had narrow abdomens in the F2 generation, how many had broad abdomens in the same generation? Show your working [2marks]
- In a related expt, fruit flies with broad abdomens were crossed with flies with narrow abdomens. The offspring with broad abdomen and the ones with narrow abdomen were in the ratio of 1:1
- What is the genotype of the parent with broad abdomen? [1mark]
- What is the biological significance of this experiment? [1mark]
- Suggest 2 reasons why fruit flies are suitable organisms to use in this genetic experiment [2marks]
- The diagram below shows a fresh water ecosystem. Study it carefully and answer the questions that follow.
- a) What is an ecosystem. [1mark]
- b) Identify the type of plants labeled S,P and Q. [3marks]
- c) Explain adaptations of plant P to their environment. [4marks]
- A healthy plant was kept in the dark for 24hours following which two of its leaves were enclosed in glass flasks as shown below. The set up was the exposed to sunlight for a number of hours.
(a) Why was it necessary to keep the plant in the dark for 24 hours? [1mark]
(b) Give the function of each of the following in the experiment
(i) Sodium hydroxide [1mark]
(ii) Sodium hydrogen corbonate [1mark]
(c) Explain the expected observations in leaf.
(i) M when tested for starch [2mark]
(ii) N when tested for starch? [2mark]
(d) Apart from light intensity, name one other aspect of light that affects photosynthesis [1mark]
SECTION B (40 MARKS)
Answer question 6 and either question 7 or 8
- An experiment was carried out to investigate a certain physiological process in plants. The experimental set-up was as follows: three vacuum flasks were labelled X, Y and Z. wet cotton wool was placed in flasks X and Y. 50 soaked bean seeds were placed in flask X; while 50 boiled and then cooled seeds were placed in flask Y. Cotton wool soaked in methylated spirit was placed in flask Z. 50 seeds, boiled, cooled and then soaked in methylated spirit was placed in flask Z and a thermometer was placed in each flask and held in place with dry cotton wool. The set-up was left standing on the side bench in the laboratory. Temperature readings were taken at the same time each day for nine days.
The results are as shown below
|
Flask |
Temperature (OC)-recorded daily |
| 1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 (DAYS) |
| X |
22 |
25 |
30 |
35 |
38 |
38 |
37 |
33 |
26 |
| Y |
20 |
20 |
20 |
22 |
25 |
30 |
33 |
39 |
45 |
| Z |
20 |
20 |
19 |
20 |
20 |
19 |
20 |
20 |
19 |
- Using the same axes plot the temperatures against time in days for flask X and Y [8marks]
- Account for the shape of the graphs from day 1 to day 9 in :
- Flask X [4marks]
- Flask Y [3marks]
- Explain the results obtained in flask Z [2marks]
- Explain why :
- Vacuum flask were used in this experiment [1mark]
- Cotton wool and not rubber bungs were used to hold the thermometer in place [1mk]
- State the aim of the above investigation [1mark]
- Evolution of an organism is towards better adaptation and survival to its habitat.
- Explain how this is achieved to members of Class Insecta towards locomotion and protection [7marks]
- Suggest how the modern long necked giraffes may have evolved from short necked stock [13marks]
- Briefly
- Describe the process of fertilization in flowering plants [16marks]
- Describe the changes that take place on the flower after fertilization [4marks]
NAME………………………………………………………………………….. ADM/NUMBER……………………….
SCHOOL………………………………………………………………… Date……………………………………
BIOLOGY (231/3)
Paper 3 (PRACTICAL)
JULY 2023
TIME: 13/4 hour
MOCKS 1 2023
Kenya Certificate of Secondary Examinations
Instructions to candidates
(a) Write your name and Admission number in the spaces provided.
(b) Answer all the questions in the spaces provided.
(c) You are required to spend the first 15 minutes of the 13/4 hours allowed for this paper reading the whole paper carefully before commencing your work.
(d) This paper consists of 6 printed pages.
(e) Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing.
For Examiner’s Use Only
| QUESTION |
MAXIMUM SCORE |
CANDIDATE SCORE |
| |
14 |
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| |
13 |
|
| |
13 |
|
| 40 |
|
- You are provided with small pieces of two tissues, labeled P and Q, obtained from an animal.
(a).Cut each specimen into two equal halves. From each specimen, crush one half and leave the other half as a solid piece. Place the solid half of specimen P into a test tube labeled K. Place the solid half of specimen Q into a test tube labeled L.
Put about 2cm3 hydrogen peroxide into each of the test tubes.
(i) State the observations made in the two test tubes. [2marks]
Test tube K ………………………………………………………………………………………
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Test tube L ………………………………………………………………………………………..
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(ii)Place the crushed specimen P into test tube labeled M and also place the crushed specimen Q into test tube labeled N. Add 2cm3 hydrogen peroxide into test tube M and N. Record the observation for each test tubes M and N in comparison to K and L [2marks]
Test tube M ……………………………………………………………………………………
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Test tube N …………………………………………………………………………………….
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(iii) Write down an equation for the reaction that was responsible for your observations in the experiments above. [1mark]
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(iv) Name the process represented by the equation in (iii) above. [1mark]
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(b) Explain how crushing affected the results of the experiments. [2marks]
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(c) Apart from the process named in (a) (iv) above, name three other functions of specimen Q[3marks]
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(d) Explain the importance of the process named in (a) (iv) above in living organisms [3marks]
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- You are provided with specimen labelled Z which has been grounded into flour.
Make a solution of the flour provided by adding water and stirring properly. Sieve or decant to obtain a solution from the mixture.
(a) (i) Using the reagents provided test for the presence of starch, proteins and lipids in the solution from specimen Z. Record the procedures, observation, and conclusions in the table below. [9marks]
| FOODSUBSTANCE |
PROCEDURE |
OBSERVATION |
CONCLUSION |
| Starch
|
|
|
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| Proteins |
|
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| Lipids |
|
|
|
(ii) From the conclusions made in (a) (i) above, suggest the regions of the alimentary canal where the digestion of specimen Z would take place. [3marks] …………………………………………………………………………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………….
(b) State one use of any two food substances found in specimen Z. [2 marks]
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- You are provided with leaves of specimens A, B, C, D, and E.
(a) Use the following features in the order in which they are listed, to prepare a dichotomous key: [8 marks]
Type of leaf
Shape of the lamina
Succulent or non-succulent
Leaf margin
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(b) (i) Name the likely habitat of specimen C. [1mark]
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(ii) Give a reason for your answer in (b) (i) above. [1mark]
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(c) State the significance of the shiny upper surface of specimen A. [2marks]
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MOCKS 1 2023
Piece of flesh labeled P
Piece of Liver labeled Q
Groundnuts flour labelled Z( each student requires about 20gms of the flour)
Leaves – (a) Broad leave with smooth margins e.g Mango labeled A
(b) Grass leaf Labeled B.
(c) Tradescantia leaf / any succulent leaf labeled C
(d) Broad leaf with serrated leaf margine.g Tobacco leaf labeled D
(e) Compound leafe.gJacaranda labeled E
Iodine solution
Absolute ethanol
10% sodium hydroxide solution
1% copper (ii) sulphate solution
Hot water bath
Means of labeling
Four test tubes
MOCKS 1 2023
- Viruses ;
- Fruits/ a part of the fruit is, fleshy/ juicy/ succulent (Fruits/ a part of the fruit is brightly coloured. Fruits large-sized and thus conspicuous (acc. Clustered rej conspicuous alone) fruits scented/ with a sweet smell/ aroma; seeds have tough/ hard/ slimy testa; (ej coat alone for testa, Accept seed coat for testa) seeds covered with sticky/ mucoid/ slimy materials (the fruit/ parts of the fruit) have hooks/ hook-like structures (mark the first floor (1 mk x 2= 2mks)
-
- Irritability
- Ability of a living thing/ organism to detect/ perceive and react to a change in its environment (note both/ the two underlined aspects of the phenomenon must be stated to score 1 mk)
- seed dormancy
- Inability of seed of a given plant to germinate (caused by conditions within the seed) in a specified period of time. Under environmental factors/ conditions that are normally suitable/ ideal o germination (of the non-dormant seed) (Note: both aspects of seed dormancy must be specified to score 1mk)
- Double fertilization
- One of the male nuclei fuses with the egg cell nucleus to form diploid zygote while the other male nucleus fuses with the polar nucleus to form triploid primary endosperm
- a ) Goitre
- b) beriberi
- increase surface area for attachment of respiratory enzymes hence increased rate of respiration
-
| Transpiration |
Guttation |
| Loss of water from a plant surface in from of water vapour |
Loss of water from a plant surface in form of liquid droplets |
| Take place/ water is lost through stomata, lenticels, thin cuticle |
Takes place/ water is lost though hydathodes; |
- pepsin secreted as pepsinogen
Trypsin secreted as trypsinogen
- cilia waft to propel the ovum in the fallopian tube
9 a) carotid artery
- b) coronary artery
- Has contractile vacuoles that collect and discharge excess water out of the cell
- a)Name taxonomic class of woodlice
- (class) Crustacea (NB initial ‘C’ must be written in upper case while others in small letters
- b) Name two other organisms in the taxonomic class in (a) (i) above
- Crab/ lobster/ catfish, Bamades, Krill
- c) State two features characteristics/ diagnostic o organisms in the taxonomic class in (a) (i) above
- Bitamoua (two pated) appendices; two body [parts; cephalothorax covered dorsally with carapace/ hard shell like structure, two pairs of antennae; Chitinous exoskeleton/ segmented body (paired and )jointed appendages/ each body segment with a pair of appendages/ dorsal, tubuler heart/ open circulatory system/ internal body cavity a haemocoel through which haemolymph circulates and in which internal body organs suspended/ vision by means of compound and simple eyes/ Excretion by means of malphigian tubules.
- a) Suggest a suitable method used to mark the grasshopper
- Applying nail polish/ varnish, using permanent marker pen/ felt pen
- b) Calculate population density of grasshoppers in the field.
Population size = first marked/ captured x second capture
Marked re-captured
= 100 x 80
40 = 200 grasshoppers
Population density = 200 grasshoppers
5km2
= 40 grasshoppers/km2
- ) Identify the cell
- ii) State 1 feature observable in the diagram above that adapts the cell to its function of absorption of water and mineral ions
- Has an elongated portion to increase surface area over which water and mineral ions are absorbed
- b) Name the part labeled F
- a) i) Identify type of neuron in the diagram above
- ii) Give one reason for your answer in (a) (i) above
- Cell body located at one end/ the edge of the cell
b)Draw an arrow alongside diagram above to indicate direction of flow of nerve impulse in the neurone.
- a) Where in the mammalian body does the reaction above take place?
- In red blood cells in the blood capillaries in the tissues with high carbon (IV) oxide concentration.
- b) Name enzyme X
- a) Name the part labeled P
- b) Give one adaptive feature of the nucleus – the organelle T.
- Carried haploid/ half the normal number of chromosomes of the species; so that following fertilization, the normal diploid condition of the species is reconstituted; rej. Contains the nucleus synthesis of ribosomes; rej. Semipermeable membrane for selective movement of materials between the nucleoplasm and the rest of the cell (NB- the core function of the nucleus in the sperm cell is reproduction. The latter are auxiliary roles of the organelle.
- c) State function of Flagellum/ tail – pat Q
- Forward- Propulsion of the cell.
- a)Why is it necessary to remove carbon (IV) oxide
- Confirm that any carbon (IV) oxide present/ detected is from the animal and not from the incoming air.
- b) Calcium hydroxide solution is put in the jars labeled 2 and 3, explain why
- Jar 2 – to confirm that the incoming air is Carbon (IV) oxide – free
- Jar 3 – to test for Carbon (IV) oxide in the air exhaled by the small animal.
- c) Suggest suitable control for this experiment
- Same set up but without/ excluding the animal
- a) i) Name the physiological process illustrated above
- ii) Give two examples of applicability of the process named in (a) (I above in plants
- Gaseous exchange/ Taking in of oxygen and excretion of carbon IV) oxide
- Translocation of materials (in phloem tissue
- Absorption/ uptake of mineral ions/ salts
- b) State 2 ways by which movement of the dye molecules would be slowed down.
- Lowering the temperature of the medium
- Increasing thickness of the membrane (to increase the diffusion distance
- Use a smaller quantity of the dye
- Add more water to reduce the concentration gradient.
- a) i) What was being investigated in the experiment?
- Positive – hydrotaxis in ants/ Response of ants to water.
- ii) Explain your answer in (a) (i) above
- Ants moved into the flask containing (cotton wool soaked in) water; to be able to obtain water for their metabolism. No ants moved into the dry flask.
- b) What was the role of flask II in the experiment?
- Outline 2adaptive features of guard cells
- Inner/ proximal/ventral wall in thicker and thus less elastic than the outer/ distal/ dorsal wall; the converse- the outer/distal/dorsal wall is thinner and thus elastic than the inner/ proximal/ ventral wall) and thus when the cells are turgid, the outer wall stretches more while the inner wall resist enlargement making the cells burg outwards leaving a pore-stoma in between.
- Contain chloroplasts and thus carry out photosynthesis; forming glucose (an osmotically active chemical substance) which increases osmotic pressure of the guard cells, water molecules them move into the cells by osmosis (resulting into opening of the stomatal pore.
- a) The scientific name of the European wildcat
- b) The species/ specific name/ epithet of the European wildcat
- silvestris; NB: all letters in this word must be lower case, to score.
- A) secretin b) cholecystokinin
- Identify type o gene mutation represented in each case
- I – Deletion (of gene ‘e’)
- II – Substitution/ inversion
- Define omnivores
- Animals that acquire energy and nutrients from materials of plant and animal origin. Omnivores are strictly animals.
- b) Name two animals that are omnivores
Man (rej men) human/ human being/, bear, hedge dog, skunk, squirrel, Raccoon, mouse, rat. Chimpazee, orangutan, Armadilo, Monkey (Accept the plural’s/ humans beings, pigs, badgers etc)
- a) Give the term used to describe this phenomenon
- Codominance (written as one word)
- b) Give one example of a trait in human beings where the condition whose term is named in (a) above express itself.
- Type AB blood group/ type, rej blood group/s alone, rej ABO blood group/s acc. Sickle cell trait rej. Sickle cell disease/ sickle cell anaemia.
- State 2 normal morphological/ physical features of the mammalian red blood cell represented in the diagram above
- Biconcave
- Lacks nucleus/ anucletae
- Ai) rate of transpiration is higher in windy than in still air condition
- ii) wind blows off water vapour from the surface of the leaf ;thus raising the saturation deficit hence increasing rate of transpiration;
- b) widenedstomatal aperture increases the surface area for transpiration
- A) increase in oxygen concentration results in corresponding increase in amount of sugar loss and potassium gain; oxygen gas is used to oxidize the sugar during respiration to generate energy needed for active uptake of potassium ions;
- b) introducing enzyme inhibitors;
reducing/eliminating oxygen gas and sugar;
- a) Identify the tissue labeled K
- Palisade/ mesophyll (tissue)
- b) Name the response (to sunlight) sown by the chloroplasts in diagram I
- Positive phototaxis/ positive phototactic response rej. Phototaxis alone, rej positive phototropism.
- Outline importance of the orientation of the chloroplasts as illustrated in diagram II i.e why do chloroplasts move away from bright sunlight?
- Chloroplasts move away from bright/high-intensity sunlight (at 2pm) to avoid solarisation / destruction/ photo –oxidation) of chlorophyll (molecules) by high – intensity/ very bright sunlight.
MOCKS 1 2023
- a ) i) rate of water absorption/uptake by the leafy shoot;
- ii) rate of transpiration;
b)(i)
Cut shoot under water;
Apply petroleum jelly to cork, glass/ bung glass
Open reservoir:
(ii)
To ensure no air enters leafy shoot xylem;
To ensure the apparatus is airtight;
To remove air bubbles from tube
(c)
Stomata size and number; Hairy leaf; Leaf size and shape
Leaf fall; Cuticle;
-
-
Population size = First capture X Second capture
Marked recaptured
= 70 X 40
27
= 104 garsshoppers
-
Large samples are more representative of the population
-
Sweep nets
-
- No organisms moved into or out of the study area during the investigation
- The grasshoppers mixed randomly and completely in the population
- Markings do not wear off
- No new hatchings / metamorphosis
- The markings do not alter the grasshoppers behaviour
-
- AA ; aa;
- 150 x 4 = 600;
¾ x 600 = 450 or 600 – 150 = 450 fruit flies ;
-
- Aa ;
- Test cross /back cross to determine an unknown genotype ;
Short life cycle
Produce many offsprings
Has clear contrasting characteristics / exist in many mutant forms
Large chromosomes
Few chromosomes
Wide spread throughout the world
Not known to be a vector of human diseases
- a) A natural unit in which abiotic and biotic factors interact freely / naturally leading to aself sustaining system
- b) S – Emergent type ;P – Submerged type; Q – Floating type ;
c)
– Have numerous chloroplasts to absorb light for photosynthesis
– Chloroplast are sensitive to low light intensities, thus photosynthesiseunder low light
Intensities
– The leaves are deeply dissected into thread like straws to increase the surface area for absorption of light
- Have a large air spaces/aerenchyma tissues for storage of air required for gaseous exchange/provide buoyancy.
- The leaf lack cuticle to facilitate exchange of gases through epidermis
- They have poorly developed roots that lack root hairs to reduce absorption
of water
5.(a) To destarch;
(b)
(i) To absorb carbon (iv) oxide gas;
(ii) To increase the concentration of carbon (iv) xide gas;
(c)
(i) M : brown colour of iodine solution is retained ;photosynthesis did not take place due to absence of carbon (iv) oxide gas hence no starch thus negative results;
(ii) N : blue-black colour was observed; all conditions necessary for photosynthesis were available hence accumulation of starch thus positive results;
(d)light duration; light wavelength;
6a)
-
- Flask X [4marks]
Steady increase in temperature reading followed by a steady decline;the starch in soaked seeds was hydrolysed by water to reducing sugars; which was oxidized to produce energy needed for germination of seeds ;after all the stored food had been used up ,the decline in energy production led to decline in temperature reading;
- Flask Y [3marks]
Steady /continuous rise in temperature reading; boiling denatured the respiratory enzymes, bacteria(decomposers) respired as they fed on boiled seeds to bring about decomposition; due to their increasing number following steady reproduction ,the heat energy produced increased exponentially;
- Explain the results obtained in flask Z [2marks]
Constant temperature reading ;Methylated spirit disinfected the seeds hence no bacteria that would have respired to release heat energy;
- Explain why :
- Vacuum flask were used in this experiment [1mark]
To conserve the heat energy produced during respiration
- Cotton wool and not rubber bungs were used to hold the thermometer in place
To allow free circulation of air in and out of the flask;. (Bacteria need oxygen gas for respiration)
- State the aim of the above investigation [1mark]
To show that germinating seeds produce energy during respiration;
To show that decomposers produce heat energy during respiration;
- Evolution of an organism is towards better adaptation and survival to its habitat.
- Explain how this is achieved to members of Class Insecta towards locomotion and protection [7mks]
Locomotion
Wings for flight
Legs for walking
Segmentation to allow movement
Moulting /ecdysis to permit locomotion
Protection
Hard cuticle for protection against mechanical injury/ infestation
Polymorphic forms / colour for camouflage against predators
Cuticle is waterproof /chitinous to prevent water loss
- Suggest how the modern long necked giraffes may have evolved from short necked stock [13mks]
Short necked stock over produced resulting to competition for food ; variation occurred in population ; long necked giraffes resulted among short necked ; both stock struggled for existence ; long necked giraffes were better adapted than short necked /fittest for survival; long necked giraffes reached reproductive age and passed on desirable trait to offspring while short necked giraffes failed to do so / were eliminated, long neck was naturally selected ; gradual changes accumulated in new stock forming a population of long necked giraffes ;
- A) process of fertilization
Fertilization is the fusion of the male and female nuclei in the embryo sac; after pollination the pollen grains absorbs nutrients from the stigma; and develops an outgrowth called the pollen tube; the pollen tube grows down the style ,to the embryo sac taking along the male gametes with it; the pollen grains usually adhere to the sigma as a result of the stigma cells secreting a sticky substance; which also stimulates the pollen grain to germinate sending down its pollen tube;the growth of the pollen tube into the stigma ,through the style to the ovary is by pushing its way between the cells where it gets nourishment from the surrounding tissues;this process is quite rapid and takes place in a matter of minutes; as the pollen grain germinates ,the tube nucleus occupies a position at the tip of the growing pollen tube; the generative nucleus divides by mitosis into two male gamete nuclei, which follow behind the tube nucleus as the pollen grows down the style; the pollen tube enters an ovule through the micropyle and when it reaches the centre of the ovule it penetrates the wall of the embryo sac and burst open; meanwhile the tube nucleus disintegrates leaving a clear way for the entry of the male nuclei; one of the male nuclei fuses with the egg cell nucleus to form a diploid zygote; which develops in to the embryo;while the other male nucleus fuses with the polar nucleus to form a triploid primary endosperm;
NAME………………………………………………………………………….. ADM/NUMBER……………………….
SCHOOL………………………………………………………………… Date……………………………………
BIOLOGY (231/3)
Paper 3 (PRACTICAL)
JULY 2023
TIME: 13/4 hour
MOCKS 1 2023
Kenya Certificate of Secondary Examinations
Instructions to candidates
(a) Write your name and Admission number in the spaces provided.
(b) Answer all the questions in the spaces provided.
(c) You are required to spend the first 15 minutes of the 13/4 hours allowed for this paper reading the whole paper carefully before commencing your work.
(d) This paper consists of 6 printed pages.
(e) Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing.
For Examiner’s Use Only
| QUESTION |
MAXIMUM SCORE |
CANDIDATE SCORE |
| |
14 |
|
| |
13 |
|
| |
13 |
|
| 40 |
|
- You are provided with small pieces of two tissues, labeled P and Q, obtained from an animal.
(a).Cut each specimen into two equal halves. From each specimen, crush one half and leave the other half as a solid piece. Place the solid half of specimen P into a test tube labeled K. Place the solid half of specimen Q into a test tube labeled L.
Put about 2cm3 hydrogen peroxide into each of the test tubes.
(i) State the observations made in the two test tubes. [2marks]
Test tube K …………less effervescence
…………………………………………………………………………………………………….
Test tube L …………more effervescence compared to K
…………………………………………………………………………………………………….
(ii)Place the crushed specimen P into test tube labeled M and also place the crushed specimen Q into test tube labeled N. Add 2cm3 hydrogen peroxide into test tube M and N. Record the observation for each test tubes M and N in comparison to K and L [2marks]
Test tube M ………more effervescence compared to K
…………………………………………………………………………………………………
Test tube N ………more effervescence compared to L
………………………………………………………………………………………………….
(iii) Write down an equation for the reaction that was responsible for your observations in the experiments above. [1mark]
Hydrogen peroxide → water + oxygen gas
(iv) Name the process represented by the equation in (iii) above. [1mark]
…………detoxification
(b) Explain how crushing affected the results of the experiments. [2marks]
Crushing increased the surface area ;upon which enzyme catalase can speed up the decomposition of toxic hydrogen peroxide ;
……………………………………………………………………………………….
(c) Apart from the process named in (a) (iv) above, name three other functions of specimen Q [3marks]
Deamination
Blood sugar regulation
Thermoregulation
Regulation of plasma proteins
Haemoglobin regulatiog
……………………………………………………………………………………….
(d) Explain the importance of the process named in (a) (iv) above in living organisms [3marks]
Harmful/ toxic substances e.g hydrogen peroxide are broken down; to less harmful substances such as water and oxygen gas; this creates a conducive environment for physiological processes in the cell;the oxygen gas produced can also be used in respiration;
- You are provided with specimen labelled Z which has been grounded into flour.
Make a solution of the flour provided by adding water and stirring properly. Sieve or decant to obtain a solution from the mixture.
(a) (i) Using the reagents provided test for the presence of starch, proteins and lipids in the solution from specimen Z. Record the procedures, observation, and conclusions in the table below. [9marks]
| FOODSUBSTANCE |
PROCEDURE |
OBSERVATION |
CONCLUSION |
| Starch
|
To 2ml of solution Z,add iodine solution dropwise;
|
Blue-black colour is observed ; |
Starch present; |
| Proteins |
To 2ml of solution Z,add equal amount of 10% sodiumhydroxide solution followed by 1% copper (ii) sulphate solution;
|
Purple/violet colour is observed ; |
Proteins present; |
| Lipids |
To 2ml of solution Z,add 4ml of absolute ethanol and shake .transfer into another test tube half filled with water;
|
Formation of white emulsion ; |
Lipids present; |
(ii) From the conclusions made in (a) (i) above, suggest the regions of the alimentary canal where the digestion of specimen Z would take place. [3marks]
Starch –mouth;duodenum;
Proteins –stomach;duodenum;
Lipids –duodenum;ileum;
(b) State one use of any two food substances found in specimen Z. [2 marks]
Starch –digested to glucose which is then oxidized during respiration to generate energy;
Proteins –digested to amino acids which are then oxidized to release energy during starvation;
- You are provided with leaves of specimens A, B, C, D, and E.
(a) Use the following features in the order in which they are listed, to prepare a dichotomous key: [8 marks]
Type of leaf
Shape of the lamina
Succulent or non-succulent
Leaf margin
1 a) leaf simple………………………………………………………….go to 2 b) leaf compound………………………………………………………E
2 a) leaf with broad lamina ……………………………………………go to 3
- b) leaf with narrow lamina……………………………………………B
3 a) leaf succulent ……………………………………………………… C
- b) leaf non-succulent……………………………………………………go to 4
4 a) leaf with serrated margin ……………………………………………..D
- b) leaf with smooth margin……………………………………………….A
………………………………………………………………………………………
(b) (i) Name the likely habitat of specimen C. [1mark]
Desert;
(ii) Give a reason for your answer in (b) (i) above. [1mark]
Succulent leaves to store water
(c) State the significance of the shiny upper surface of specimen A. [2marks]
Shiny to reflect light away; hence reduce the rate of transpiration;
